Answer:
Explanation:
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An alternating current is supplied to an electronic component with a rating that it be used only for voltages below 12 V. What is the highest Vrms that can be supplied to this component while staying below the voltage limit
Answer:
[tex]V_{rms}=6\sqrt2\ V[/tex]
Explanation:
Given that,
The maximum voltage of an alternating current, [tex]V_{max}=12\ V[/tex]
We need to find the highest Vrms that can be supplied to this component while staying below the voltage limit.
Let rms voltage in terms of peak voltage is given by :
[tex]V_{rms}=\dfrac{V}{\sqrt2}\\\\=\dfrac{12}{\sqrt2}\\\\=\dfrac{12}{\sqrt2}\times \dfrac{\sqrt2}{\sqrt2}\\\\=\dfrac{12\sqrt2}{2}\\\\=6\sqrt2\ V[/tex]
Hence, the required rms voltage is [tex]6\sqrt2\ V[/tex].
A fluid flows though a horizontal 0.1 inch diameter pipe. When the Reynold number is 1508, the head loss over a 20-ft length of the pipe is 6.2 ft. Determine the fluid velocity.
Answer:
The fluid velocity V = 1.98 ft/s
Explanation:
From the information given:
The fluid velocity can be determined from the head-loss [tex]h_L[/tex] of a laminar pipe and it is expressed as:
[tex]h_L = f\dfrac{l\times V^2}{D \times 2g}[/tex]
where;
f = frictional factor ; l = length; D = diameter; V= fluid velocity and g = acceleration due to gravity.
And;
[tex]f = \dfrac{64}{Re}[/tex]
For fluid movement in a laminar flow, the Reynolds number (Re) is usually lesser than 2100.
Given that:
Re = 1508 < 2100 ( laminar flow)
Then;
[tex]f = \dfrac{64}{1508}[/tex]
f = 0.04244
Also;
the head-loss [tex]h_L[/tex] = 6.2 ft
frictional force f = 0.04244
length = 20-ft
acceleration due to gravity (g) = 32.2 ft/s²
Replacing all the values into the equation [tex]h_L = f\dfrac{l\times V^2}{D \times 2g}[/tex]; the fluid velocity is:
[tex]6.2 = 0.04244 \times \dfrac{20 \times V^2}{0.1 \times \dfrac{1}{12} \times 2\times 32.2}[/tex]
[tex]6.2 = 0.04244 \times \dfrac{20 \times V^2}{0.53667}[/tex]
6.2 × 0.53667 = 0.04244 × 20 × V²
3.327354 = 0.8488 × V²
[tex]V^2= \dfrac{3.327354} { 0.8488}[/tex]
[tex]V^2=3.92[/tex]
[tex]V = \sqrt{3.92}[/tex]
V = 1.98 ft/s
Water vapor at a rate of 40,000 kg/h and at 8 MPa and 500 C enters an adiabatic turbine and leaves the turbine at 40 kPa as saturated vapor. The rate of entropy generation for this turbine is close to,
Answer:
M(dot)= 40000 kg/h Pressure= 8Mpa and T= 500C If its adiabatic then you know that it loses no heat or Q(dot) =0 You have a turbine which drives a shaft producing work for the other systems.We want to find entropy production of alpha.Type the correct answer in the box. Spell all words correctly.
A genetically engineered hormone, , can treat Mary’s child for growth hormone deficiencies by stimulating body growth and increasing muscle mass.
Answer:
recombinant human growth
Explanation:
Answer:
recombinant human growth
Explanation:
yes
A piston–cylinder device contains 15 kg of saturated refrigerant-134a vapor at 280kPa. A resistance heater inside the cylinder with a value of 110V is used to transfer heat of 20.67kJ/kg to the refrigerant at constant pressure for 6min. Determine the current supplied if the final temperature is 75℃
Answer:
18.77 A
Explanation:
To solve this we use the energy balance equation, that is:
[tex]E_{in}-E_{out}=\Delta E_{sys}\\\\Q_{in}+W_{in}+W_{out}=\Delta U\\\\Q_{in}+W_{in}=\Delta U\\\\But\ W_{in}=Voltage(V)*Current(I)*change\ in\ time(\Delta t)=VI\Delta t,\Delta U=m(h_2-h_1)\\\\Given\ that\ m=15kg,V=110\ V,\Delta t=6\ min = (6*60\ s)=360\ s,Q_{in}=20.67\ kJ/kg*15\ kg=310.05\ kJ=310050\ J\\\\From\ table: At\ P_1=280kPa, h_1=249.71\ kJ/kg=249710\ J/kg;At\ P_2=280kPa \ and\ T_1=75^oC,P_2=319.95\ kJ/kg=319950\ J/kg\\\\Substituting:\\\\310500+(110*360*I)=15(319950-249710)\\\\[/tex]
[tex]39600I=1053600-310500\\\\39600I=743100\\\\I=18.77\ A[/tex]
who designs each page or screen with which a user interacts?
Answer:
UI Design
Explanation:
